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Math Help - definite integral, evaluation

  1. #1
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    definite integral, evaluation

    please se attachmnt for question:
    Attached Thumbnails Attached Thumbnails definite integral, evaluation-maths-question-5.bmp  
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  2. #2
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    If you know the result of this famous integral :

     \int_0^{\frac{\pi}{2} } \ln[\sin(x)]~dx =  -\frac{\pi}{2}\ln{2}

    what to do with your problem is changing  1 + \cos(x) to  2 \cos^2(\frac{x}{2}) , and remind that

     \int_0^{\pi } \ln[\cos(\frac{x}{2})] ~dx = 2 \int_0^{\frac{\pi}{2} } \ln[\cos(x)]~dx  = 2 \int_0^{\frac{\pi}{2} } \ln[\sin(x)]~dx  =  -\pi \ln{2}


    Or another way :

    Consider

     \int_0^{\pi} = \int_0^{\frac{\pi}{2}} + \int_{\frac{\pi}{2}}^{\pi}

     \int_0^{\pi} \ln[ 1 + \cos(x) ] ~dx = \int_0^{\frac{\pi}{2}} \ln[ 1 + \cos(x) ] ~dx   + \int_{\frac{\pi}{2}}^{\pi} \ln[ 1 + \cos(x) ] ~dx

    for the second integral , sub .  t = \pi - x we obtain

     \int_0^{\frac{\pi}{2}} \left( \ln[ 1 + \cos(x) ]  + \ln[ 1 - \cos(x) ]   \right ) ~dx

     = 2 \int_0^{\frac{\pi}{2}} \ln[\sin(x)]~dx = -\pi\ln{2}


    Edited :

    I just find another solution by means of magic differentiation :

    Let

     I(a) = \int_L \ln[ 1 + a \cos(x) ] ~dx

    for brevity , I replace \int_0^{\frac{\pi}{2}} with  \int_L

    Take differentiation ,

     I'(a) = \int_L \frac{ \cos(x) }{1 + a \cos(x) } ~dx

     I'(a) = \int_L [ \frac{1}{a} - \frac{1}{a} \cdot \frac{1}{1 + a \cos(x)}] ~dx

     = \frac{\pi}{a} - \frac{\pi}{a} \cdot \frac{1}{\sqrt{1 - a^2}}

    Therefore ,

     I(a)  = \pi \left ( \ln(a) - \int \frac{1}{a\sqrt{1 - a^2 }} ~da \right )

    Sub.  a = 1/t

     I(a) = \pi \left ( \ln(a) - \ln( a + \sqrt{1 - a^2} ) + \ln(a)\right ) + K

     = \pi \ln(a + \sqrt{1-a^2})+K

    we know  I = I(1) - I(0) so

     I = -\pi\ln{2}
    Last edited by simplependulum; March 2nd 2010 at 11:41 PM.
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  3. #3
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    here's a completely different approach:

    \int_{0}^{\pi }{\ln (1+\cos x)\,dx}=\int_{0}^{\pi }{\int_{0}^{1}{\frac{\cos x}{1+t\cos x}\,dt}\,dx}, reverse integration order and after some straightforward computations we get \pi \int_{0}^{1}{\frac{\sqrt{1-t^{2}}-1}{t\sqrt{1-t^{2}}}\,dt}.

    substitution u=\ln\big(1+\sqrt{1-t^2}\big) solves the integral achieving the value -\ln2, thus the original integral equals -\pi\ln2, as shown above.
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